The Discrete Fourier Transform CS/BIOEN 4640: Image Processing - PowerPoint PPT Presentation

The Discrete Fourier Transform CS/BIOEN 4640: Image Processing Basics March 27, 2012

Review: Fourier Transform Given a complex-valued function g : R → C , Fourier transform produces a function of frequency ω : � ∞ 1 � � √ G ( ω ) = g ( x ) · cos ( ω x ) − i · sin ( ω x ) dx 2 π −∞ � ∞ 1 g ( x ) · e − i ω x dx = √ 2 π −∞

Review: The Dirac Delta Definition The Dirac delta or impulse is defined as � ∞ δ ( x ) = 0 for x � = 0 , δ ( x ) dx = 1 and −∞ ◮ The Dirac delta is not a function ◮ It is undefined at x = 0 . ◮ Has the property � ∞ f ( x ) δ ( x ) dx = f ( 0 ) for any function f −∞

The Fourier Transform of a Dirac Delta � ∞ 1 δ ( x ) · e − i ω x dx F{ δ ( x ) } ( ω ) = √ 2 π −∞ 1 e 0 √ = 2 π 1 = √ 2 π ◮ In other words, Fourier of a Dirac is constant ◮ So, it has equal response at all frequencies

Convolution with a Dirac Delta Convolving a function g ( x ) with a Dirac delta gives � ∞ ( g ∗ δ )( x ) = g ( y ) · δ ( y − x ) dy −∞ = g ( x ) ◮ So, convolving with Dirac is the identity operator ◮ Also can be seen in the Fourier domain: √ F{ g ∗ δ } = 2 π F{ g } · F{ δ } = F{ g }

The Comb Definition The comb function or Shah function is defined as an infinite sum of Dirac deltas: ∞ � III ( x ) = δ ( x − k ) k = −∞ 1.5 ● ● ● ● ● 1.0 ● ● ● ● ● ● Notice that just like the III(x) 0.5 Dirac delta, the comb function is not a function −5 −4 −3 −2 −1 0 1 2 3 4 5 −0.5

Using the Comb to Sample Given a continuous function g ( x ) , we can “sample” this function by multiplication by a comb: ¯ g ( x ) = g ( x ) · III ( x ) ∞ � = g ( k ) · δ ( x − k ) k = −∞ Notice that ¯ g ( x ) is also not a function.

Using the Comb to Sample

The Discrete Fourier Transform For a discrete signal g ( u ) , where u = 0 , 1 , . . . , M , the discrete Fourier transform is given by M − 1 1 2 π mu 2 π mu � � � � �� � √ G ( m ) = g ( u ) · − i · sin cos M M M u = 0 M − 1 1 � g ( u ) · e − i 2 π mu √ = M M u = 0

Comparing Discrete and Continuous Continuous Fourier Transform: � ∞ 1 g ( x ) · e − i ω x dx G ( ω ) = √ 2 π −∞ Discrete Fourier Transform: M − 1 1 g ( u ) · e − i 2 π mu � G ( m ) = √ M M u = 0

Inverse Discrete Fourier Transform The inverse DFT, analagous to the continuous case, just changes the sign in the exponent: M − 1 1 � G ( m ) · e i 2 π mu g ( u ) = √ M M u = 0

The Fourier Transform of the Comb Fourier transform of a comb is another comb: � ω � F { III ( x ) } = III 2 π

The Fourier Transform of the Comb Spacing of comb in time domain is inversely related to spacing in frequency domain: � τω � x � �� � F = τ III III τ 2 π

Sampled Signals in the Fourier Domain If ¯ g ( x ) = g ( x ) · III ( x ) , then � ω √ � ¯ G ( ω ) = 2 π G ( ω ) ∗ III 2 π ◮ We know convolution with δ is the identity ◮ So, this produces copies of the spectrum G shifted to each peak of the comb.

Aliasing Caused by Sampling

The Nyquist-Shannon Sampling Theorem Theorem A bandlimited continuous signal can be perfectly reconstructed from a set of uniformly-spaced samples if the sampling frequency is twice the bandwidth of the signal.